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a) Ta có: \(x^2+9x+20\)

\(=x^2+4x+5x+20\)

\(=x\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(x+5\right)\)

b) Ta có: \(x^2+x-12\)

\(=x^2+4x-3x-12\)

\(=x\left(x+4\right)-3\left(x+4\right)\)

\(=\left(x+4\right)\left(x-3\right)\)

c) Ta có: \(6x^2-11x-16\)

\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)

\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)

\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)

\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)

d) Ta có: \(4x^2-8x-5\)

\(=4x^2-10x+2x-5\)

\(=2x\left(2x-5\right)+\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x+1\right)\)

e) Ta có: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)

\(=\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]\)

\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)

g) Ta có: \(x^3+9x^2+23x+15\)

\(=x^3+x^2+8x^2+8x+15x+15\)

\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+8x+15\right)\)

\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)

\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)

\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

h) Ta có: \(2x^4-x^3-9x^2+13x\)

\(=x\left(2x^3-x^2-9x+13\right)\)

i) Ta có: \(x^4+2x^3-16x^2-2x+15\)

\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)

\(=x^3\left(x-3\right)+5x^2\left(x-3\right)-x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(x^3+5x^2-x-5\right)\)

\(=\left(x-3\right)\left[x^2\left(x+5\right)-\left(x+5\right)\right]\)

\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)

\(=\left(x-3\right)\left(x+5\right)\left(x-1\right)\left(x+1\right)\)

8 tháng 7 2016

b, \(\left(x^2+x\right)^2+4x^2+4x-12=x^4+2x^3+x^2+4x^2+4x-12\)

                                                         \(=x^4+2x^3+5x^2+4x-12\)

                                                         \(=\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)\)

                                                         \(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)

                                                          \(=\left(x^3+3x^2+8x+12\right)\left(x-1\right)\)

                                                          \(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\left(x-1\right)\)

                                                           \(=\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)\)

                                                            \(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)

c,        \(x^3+3x^2-4=\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(2x+4\right)\)

                                    \(=x^2\left(x+2\right)+x\left(x+2\right)-2\left(x+2\right)\)

                                     = \(\left(x^2+x-2\right)\left(x+2\right)\)

9 tháng 7 2016

a)\(x^5+x^4+1=x^5-\left(-x^3+x^3\right)+x^4+\left(x^2-x^2\right)+\left(x-x\right)+1\)

\(=x^5-x^3+x^2+x^4-x^2+x+x^3-x+1\)

\(=x^2\left(x^3-x+1\right)+x\left(x^3-x+1\right)+\left(x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)

b,c có ng lm rồi

d)\(2x^4-3x^3-7x^2+6x+8\)

Ta thấy x=-1 là nghiệm của đa thức 

=>đa thức có 1 hạng tử là x+1

\(\Rightarrow\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)

\(\Rightarrow\left(x+1\right)\left[2x^3-x^2-4x-4x^2+2x+8\right]\)

\(\Rightarrow\left(x+1\right)\left[x\left(2x^2-x-4\right)-2\left(2x^2-x-4\right)\right]\)

\(\Rightarrow\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)

phần còn lại bạn tự lo nhé

8 tháng 10 2016

đề bài ???

8 tháng 10 2016

c)4x^4x^− x^− x = x*(4x^3 + 4x^2 - x -1)

21 tháng 10 2018

a) \(x^2-6x+9-9y^2=x^2-2\cdot3+3^2-\left(3y\right)^2=\left(x-3\right)^2-\left(3y\right)^2=\left(x-3-3y\right)\cdot\left(x-3+3y\right)\)

21 tháng 10 2018

b) \(x^3-3x^2+2x-1+2\cdot\left(x^2-x\right)=\left(x-1\right)^3+2x\cdot\left(x-1\right)=\left(x-1\right)\cdot\left[\left(x-1\right)^2+2x\right]\)

3 tháng 6 2017

a)\(x^2+9x+20\)

\(\Leftrightarrow x^2+4x+5x+20\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)

\(\Leftrightarrow\left(x+5\right)\left(x+4\right)\)

b)\(x^2+x-12\)

\(\Leftrightarrow x-3x+4x-12\)

\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\)

3 tháng 6 2017

Vừa vừa phải phải thôi người ta mất công gửi lên còn chửi người ta đó điên mất lịch sự

21 tháng 7 2021

Trả lời:

a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)

d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)

e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)